Typically, copper busbars are used when high currents at (relatively) low frequency need to be delivered to some load. Their use is favoured maily by the possibility of having a large conductor section and/or lower stray capacitance compared to an insulated cable. Typically insulation is ensured by air in these cases. The use in the 50/60 Hz AC range is acceptable due to skin depth being about 9 mm @60 Hz.
During the design phase of an electrical system using cables and/or busbars carrying high currents, it may be of interest, for mechanical reasons, to calculate the maximum force acting between two busbars placed near each other. Indeed, when two conductors carry current, due to Lorentz's force, they experience a force acting on them. If we consider two raw copper cables parallel to each other on a given plane and carrying a constant current I, the force acting on them is equal to:
$$F' = \frac{\mu_0}{4 \pi}\frac{I^2}{d}$$
Where
- $F'$ is the force per unit length [N/m]
- $\mu_0$ is the magnetic permeability of vacuum [H/m]
- $I$ is the current flowing in each conductor [A]
- $d$ is the distance between the conductors [m]
In the general case, the conductor will not necessarily have a circular section, such as in the case of busbars. As a consequence, a corrective coefficient $k$ can be used to take into account the different geometry and obtaining the following formula
$$F' = k \frac{\mu_0}{4 \pi}\frac{I^2}{d}$$
This corrective coefficient depends on the geometry of the conductors and in the case of rectangular section copper bars can be obtained from Dwight's chart
This chart (or similar ones) are usually found in application notes given by manufacturers.
The combined use of the formula and the chart allows to calculate the maximum electrodynamic stress that could occur in the system. However, if a better picture of the system is needed, finite element simulation can be used to get an even better picture. In the following, both paths are illustrated.
Consider, for instance, the case presented in the picture below
In this case we have
- A peak current of 800 A in each copper busbar
- Width (a) of 6 mm
- Height (b) of 60 mm
- Depth (L) into the page of 1000 mm
- A distance between the centers of the copper busbars of 12 mm
- b/a = 10
- d/a = 2
If a finite element simulation is performed using FEMM, a very close result of 4.63 N/m is obtained.
The result can be obtained by simulating the steady state magnetic field when 800 A circulate on each busbar and then integrating J and B over the busbar surface (note that FEMM gives you a result in N, but by setting a depth of 1000 mm it is possible to obtain the force per unit meter, alternatively you can simply divide by the depth of the busbars).
It may also be of interest to check out the H field from one busbar center to the other (the B field is the same since copper and air have a similar behaviour when it comes to magnetic permeability and B is simply proportional to H through $\mu_0$).
This example was reproduced from [1].
If you are curious and want to learn more on the topic, I suggest you these two readings
[1] M. C. Petrescu, L. Petrescu, Electrodynamic forces between two DC busbars distribution systems conductors, 2016, U.P.B. Sci. Bull., Series C, Vol. 78, Issue 2.
[2] J.P. Thierry, C. Kilindjian, Electrodynamic forces on busbars in LV systems, 1996, E/CT, Cahier Technique Merlin Gerin n° 162/p2.
